Question

1) What is the percent by mass of water in the hydrate?
2) What is the percent by mass of the anhydrous salt in the hydrate?

Answer 1

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1] What is the percent by mass of water in the hydrate?

2] What is the percent by mass of the anhydrous salt in the hydrate?

$\huge\mathbb\red{Answer:-}$

1]The theoretical (actual) percent hydration (percent water) can be calculated from the formula of the hydrate by dividing the mass of water in one mole of the hydrate by the molar mass of the hydrate and multiplying this fraction by 100.

2]Experimentally measuring the percent water in a hydrate involves first heating a known mass of the hydrate to remove the waters of hydration and then measuring the mass of the anhydrate remaining. The difference between the two masses is the mass of water lost. Dividing the mass of the water lost by the original mass of hydrate used is equal to the fraction of water in the compound. Multiplying this fraction by 100 gives the percent water.

Answer 2

Answer:

1.36%

A mole of CuSO4•5H2O contains 5 moles of water (which corresponds to 90 grams of water) as part of its structure. Therefore, the substance CuSO4•5H2O always consists of 90/250 or 36% water by weight. The water is a definite part of the structure of the compound, and the substance should not be considered to be “wet.”

2.Often, a hydrated salt can be converted into an anhydrous salt (one without water) by strong heating to drive off the water molecules. The mass of the hydrate will differ from the mass of the anhydrous salt by the mass of water that was removed during heating.

Explanation:

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